Find the limit as $x$ approaches positive infinity. $\lim_{x\to\infty}\dfrac{5x^2+4}{\sqrt{4x^4+5x}}=$
Explanation: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x^2$, let's divide by $x^2$. In the denominator, let's divide by $\sqrt{x^4}$, since for any value, $x^2=\sqrt{x^4}$. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{5x^2+4}{\sqrt{4x^4+5x}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{5x^2+4}{x^2}}{\dfrac{\sqrt{4x^4+5x}}{\sqrt{x^4}}} \gray{\text{Divide sides by }x^2=\sqrt{x^4}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to\infty}\dfrac{\dfrac{5\cancel{x^2}}{\cancel{x^2}}+\dfrac{4}{x^2}}{\sqrt{\dfrac{4\cancel{x^4}}{\cancel{x^4}}+\dfrac{5\cancel{x}}{\cancel{x}\cdot x^3}}} \\\\ &=\lim_{x\to\infty}\dfrac{5+\dfrac{4}{x^2}}{\sqrt{4+\dfrac{5}{x^3}}} \\\\ &=\lim_{x\to\infty}\dfrac{5+0}{\sqrt{4+0}} \gray{\lim_{x\to\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{5}{\sqrt{4}} \\\\ &=\dfrac{5}{2} \end{aligned}$ In conclusion, $\lim_{x\to\infty}\dfrac{5x^2+4}{\sqrt{4x^4+5x}}=\dfrac{5}{2}$.